Problem: Solve for $x$ : $(x + 3)^2 - 16 = 0$
Add $16$ to both sides so we can start isolating $x$ on the left: $ (x + 3)^2 = 16$ Take the square root of both sides to get rid of the exponent. $ \sqrt{(x + 3)^2} = \pm \sqrt{16}$ Be sure to consider both positive and negative $4$ , since squaring either one results in $16$ $ x + 3 = \pm 4$ Subtract $3$ from both sides to isolate $x$ on the left: $ x = -3 \pm 4$ Add and subtract $4$ to find the two possible solutions: $ x = 1 \text{or} x = -7$